Algebra Examples

$f (x) = x^{6} - 2 x^{4} - 5 x^{2} + 6$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2, \pm 3, \pm 6$

$q = \pm 1$

Step 2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2, \pm 3, \pm 6$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(1)}^{6} - 2 {(1)}^{4} - 5 {(1)}^{2} + 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

One to any power is one.

$1 - 2 {(1)}^{4} - 5 {(1)}^{2} + 6$

Step 4.1.2

One to any power is one.

$1 - 2 \cdot 1 - 5 {(1)}^{2} + 6$

Step 4.1.3

Multiply $- 2$ by $1$ .

$1 - 2 - 5 {(1)}^{2} + 6$

Step 4.1.4

One to any power is one.

$1 - 2 - 5 \cdot 1 + 6$

Step 4.1.5

Multiply $- 5$ by $1$ .

$1 - 2 - 5 + 6$

$1 - 2 - 5 + 6$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $2$ from $1$ .

$- 1 - 5 + 6$

Step 4.2.2

Subtract $5$ from $- 1$ .

$- 6 + 6$

Step 4.2.3

Add $- 6$ and $6$ .

$0$

$0$

$0$

Step 5

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{6} - 2 x^{4} - 5 x^{2} + 6}{x - 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$

Step 6.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(0)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$
	$1$

Step 6.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$
	$1$	$1$

Step 6.5

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(- 2)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$
	$1$	$1$

Step 6.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$
	$1$	$1$	$- 1$

Step 6.7

Multiply the newest entry in the result $(- 1)$ by the divisor $(1)$ and place the result of $(- 1)$ under the next term in the dividend $(0)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$
	$1$	$1$	$- 1$

Step 6.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$
	$1$	$1$	$- 1$	$- 1$

Step 6.9

Multiply the newest entry in the result $(- 1)$ by the divisor $(1)$ and place the result of $(- 1)$ under the next term in the dividend $(- 5)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$
	$1$	$1$	$- 1$	$- 1$

Step 6.10

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$
	$1$	$1$	$- 1$	$- 1$	$- 6$

Step 6.11

Multiply the newest entry in the result $(- 6)$ by the divisor $(1)$ and place the result of $(- 6)$ under the next term in the dividend $(0)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$	$- 6$
	$1$	$1$	$- 1$	$- 1$	$- 6$

Step 6.12

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$	$- 6$
	$1$	$1$	$- 1$	$- 1$	$- 6$	$- 6$

Step 6.13

Multiply the newest entry in the result $(- 6)$ by the divisor $(1)$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$	$- 6$	$- 6$
	$1$	$1$	$- 1$	$- 1$	$- 6$	$- 6$

Step 6.14

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$0$	$- 2$	$0$	$- 5$	$0$	$6$
		$1$	$1$	$- 1$	$- 1$	$- 6$	$- 6$
	$1$	$1$	$- 1$	$- 1$	$- 6$	$- 6$	$0$

Step 6.15

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{5} + 1 x^{4} - 1 x^{3} + - 1 x^{2} + (- 6) x - 6$

Step 6.16

Simplify the quotient polynomial.

$x^{5} + x^{4} - x^{3} - x^{2} - 6 x - 6$

$x^{5} + x^{4} - x^{3} - x^{2} - 6 x - 6$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Regroup terms.

$x^{5} - x^{3} - 6 x + x^{4} - x^{2} - 6 = 0$

Step 7.1.2

Factor $x$ out of $x^{5} - x^{3} - 6 x$ .

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Step 7.1.2.1

Factor $x$ out of $x^{5}$ .

$x \cdot x^{4} - x^{3} - 6 x + x^{4} - x^{2} - 6 = 0$

Step 7.1.2.2

Factor $x$ out of $- x^{3}$ .

$x \cdot x^{4} + x (- x^{2}) - 6 x + x^{4} - x^{2} - 6 = 0$

Step 7.1.2.3

Factor $x$ out of $- 6 x$ .

$x \cdot x^{4} + x (- x^{2}) + x \cdot - 6 + x^{4} - x^{2} - 6 = 0$

Step 7.1.2.4

Factor $x$ out of $x \cdot x^{4} + x (- x^{2})$ .

$x (x^{4} - x^{2}) + x \cdot - 6 + x^{4} - x^{2} - 6 = 0$

Step 7.1.2.5

Factor $x$ out of $x (x^{4} - x^{2}) + x \cdot - 6$ .

$x (x^{4} - x^{2} - 6) + x^{4} - x^{2} - 6 = 0$

$x (x^{4} - x^{2} - 6) + x^{4} - x^{2} - 6 = 0$

Step 7.1.3

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x ({(x^{2})}^{2} - x^{2} - 6) + x^{4} - x^{2} - 6 = 0$

Step 7.1.4

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (u^{2} - u - 6) + x^{4} - x^{2} - 6 = 0$

Step 7.1.5

Factor $u^{2} - u - 6$ using the AC method.

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Step 7.1.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 7.1.5.2

Write the factored form using these integers.

$x ((u - 3) (u + 2)) + x^{4} - x^{2} - 6 = 0$

$x ((u - 3) (u + 2)) + x^{4} - x^{2} - 6 = 0$

Step 7.1.6

Factor.

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Step 7.1.6.1

Replace all occurrences of $u$ with $x^{2}$ .

$x ((x^{2} - 3) (x^{2} + 2)) + x^{4} - x^{2} - 6 = 0$

Step 7.1.6.2

Remove unnecessary parentheses.

$x (x^{2} - 3) (x^{2} + 2) + x^{4} - x^{2} - 6 = 0$

$x (x^{2} - 3) (x^{2} + 2) + x^{4} - x^{2} - 6 = 0$

Step 7.1.7

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x (x^{2} - 3) (x^{2} + 2) + {(x^{2})}^{2} - x^{2} - 6 = 0$

Step 7.1.8

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (x^{2} - 3) (x^{2} + 2) + u^{2} - u - 6 = 0$

Step 7.1.9

Factor $u^{2} - u - 6$ using the AC method.

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Step 7.1.9.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Step 7.1.9.2

Write the factored form using these integers.

$x (x^{2} - 3) (x^{2} + 2) + (u - 3) (u + 2) = 0$

$x (x^{2} - 3) (x^{2} + 2) + (u - 3) (u + 2) = 0$

Step 7.1.10

Replace all occurrences of $u$ with $x^{2}$ .

$x (x^{2} - 3) (x^{2} + 2) + (x^{2} - 3) (x^{2} + 2) = 0$

Step 7.1.11

Factor $(x^{2} - 3) (x^{2} + 2)$ out of $x (x^{2} - 3) (x^{2} + 2) + (x^{2} - 3) (x^{2} + 2)$ .

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Step 7.1.11.1

Factor $(x^{2} - 3) (x^{2} + 2)$ out of $x (x^{2} - 3) (x^{2} + 2)$ .

$(x^{2} - 3) (x^{2} + 2) (x) + (x^{2} - 3) (x^{2} + 2) = 0$

Step 7.1.11.2

Factor $(x^{2} - 3) (x^{2} + 2)$ out of $(x^{2} - 3) (x^{2} + 2)$ .

$(x^{2} - 3) (x^{2} + 2) (x) + (x^{2} - 3) (x^{2} + 2) (1) = 0$

Step 7.1.11.3

Factor $(x^{2} - 3) (x^{2} + 2)$ out of $(x^{2} - 3) (x^{2} + 2) (x) + (x^{2} - 3) (x^{2} + 2) (1)$ .

$(x^{2} - 3) (x^{2} + 2) (x + 1) = 0$

$(x^{2} - 3) (x^{2} + 2) (x + 1) = 0$

$(x^{2} - 3) (x^{2} + 2) (x + 1) = 0$

Step 7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} - 3 = 0$

$x^{2} + 2 = 0$

$x + 1 = 0$

Step 7.3

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

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Step 7.3.1

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Step 7.3.2

Solve $x^{2} - 3 = 0$ for $x$ .

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Step 7.3.2.1

Add $3$ to both sides of the equation.

$x^{2} = 3$

Step 7.3.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{3}$

Step 7.3.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 7.3.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{3}$

Step 7.3.2.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{3}$

Step 7.3.2.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{3}, - \sqrt{3}$

$x = \sqrt{3}, - \sqrt{3}$

$x = \sqrt{3}, - \sqrt{3}$

$x = \sqrt{3}, - \sqrt{3}$

Step 7.4

Set $x^{2} + 2$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set $x^{2} + 2$ equal to $0$ .

$x^{2} + 2 = 0$

Step 7.4.2

Solve $x^{2} + 2 = 0$ for $x$ .

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Step 7.4.2.1

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

Step 7.4.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 2}$

Step 7.4.2.3

Simplify $\pm \sqrt{- 2}$ .

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Step 7.4.2.3.1

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Step 7.4.2.3.2

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{2}$

Step 7.4.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{2}$

$x = \pm i \sqrt{2}$

Step 7.4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 7.4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i \sqrt{2}$

Step 7.4.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i \sqrt{2}$

Step 7.4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i \sqrt{2}, - i \sqrt{2}$

$x = i \sqrt{2}, - i \sqrt{2}$

$x = i \sqrt{2}, - i \sqrt{2}$

$x = i \sqrt{2}, - i \sqrt{2}$

Step 7.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 7.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 7.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

$x = - 1$

Step 7.6

The final solution is all the values that make $(x^{2} - 3) (x^{2} + 2) (x + 1) = 0$ true.

$x = \sqrt{3}, - \sqrt{3}, i \sqrt{2}, - i \sqrt{2}, - 1$

$x = \sqrt{3}, - \sqrt{3}, i \sqrt{2}, - i \sqrt{2}, - 1$

Step 8

The polynomial can be written as a set of linear factors.

$(x - 1) (x - \sqrt{3}) (x + \sqrt{3}) (x - i \sqrt{2}) (x + \sqrt{2} i) (x + 1)$

Step 9

These are the roots (zeros) of the polynomial $x^{6} - 2 x^{4} - 5 x^{2} + 6$ .

$x = 1, \sqrt{3}, - \sqrt{3}, i \sqrt{2}, - i \sqrt{2}, - 1$

Step 10

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$